3a.
$A=(x-4y)^2=(-3)^2=9$
3b.
$B=(9x^2+4y^2+12xy)-2023=(3x+2y)^2-2023$
$=50^2-2023=2500-2023=477$
3c.
$C=(x-3y)^2-(x^2-4y^2)=(2+3)^2-(2^2-4)$
$=5^2=25$
3d.
$D=(x+2y)^3=(-2y+2y)^3=0$
3e.
$(x-2)^2+y^2=0$
Mà $(x-2)^2\geq 0; y^2\geq 0$ với mọi $x,y$. Do đó để tổng của chúng bằng $0$ thì $x-2=y=0$
$\Rightarrow x=2; y=0$.
Khi đó:
$F=-(2.2-0)^3-2(2.2-0)^2-0^3$
$=-4^3-2.4^2=-96$
3g.
$x+y=2\Rightarrow x=2-y=2-(-3)=5$
$G=(x^3+y^3)+3(8x^3-y^3)$
$=x^3+y^3+24x^3-3y^3=25x^3-2y^3$
$=25.5^3-2.(-3)^3=3179$
3h.
$y=3x-5=3.2-5=1$
$H=x^3+27y^3+27x^3-y^3=28x^3+26y^3$
$=28.2^3+26.1^3=250$
Bài 4:
a. $A=4-x^2-(x+3)^2=4-5^2-8^2=-85$
b. $B=(4x^2+20x+25)-4(x^2-9)$
$=20x+61=20.\frac{1}{10}+61=2+61=63$
c.
$C=(x^3-3x^2+3x-1)+2024=(x-1)^3+2024$
$=(101-1)^3+2024=100^3+2024=1002024$
d.
$D=(x^3-6x^2+12x-8)-92=(x-2)^3-92$
$=(-98-2)^3-92=(-100)^3-92=-1000092$
e.
$E=(x+1)^3+6(x+1)^3+12(x+1)+8$
$=(x+1+2)^3=(x+3)^3=(5+3)^3=512$
